Area of a disk

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The region inside a circle, a disk, has an area of πr² when the circle has radius r. Here the symbol “π” denotes, as usual, the constant pi, defined as the ratio of the circumference of a circle to the diameter.

Modern mathematics commonly obtains the area using the methods of integral calculus, or its more sophisticated offspring, real analysis. However, in Ancient Greece the great mathematician Archimedes used the tools of Euclidean geometry to show that the area inside a circle is equal to that of a right triangle whose base has the length of the circle's circumference and whose height equals the circle's radius. The circumference is 2πr, and the area of a triangle is half the base times the height, so we reach the same conclusion.

1 Archimedes proof
- 1.1 Not greater
- 1.2 Not less
2 Reassembly proof
3 Onion proof
4 Fast approximation
- 4.1 Derivation
5 Dart approximation
6 Finite rearrangement
7 Generalizations
8 References
9 External links

[edit] Archimedes proof

If the area of the circle is not equal to that of the triangle, then it must be either greater or less. We eliminate each of these by contradiction, leaving equality as the only possibility. We use regular polygons in an essential way.

[edit] Not greater

Circle with square and octagon inscribed, showing area gap

Suppose the circle area, C, may be greater than the triangle area, T = ¹⁄₂cr. Let E denote the excess amount. Inscribe a square in the circle, so that its four corners lie on the circle. Between the square and the circle are four segments. If the total area of those gaps, G₄, is greater than E, split each arc in half. This makes the inscribed square into an inscribed octagon, and produces eight segments with a smaller total gap, G₈. Continue splitting until the total gap area, G_n, is less than E. Now the area of the inscribed polygon, P_n = C−G_n, must be greater than that of the triangle.

$\begin{align} E &{}= C - T \\ &{}> G_n \\ P_n &{}= C - G_n \\ &{}> C - E \\ P_n &{}> T \end{align}$

But this forces a contradiction, as follows. Draw a perpendicular from the center to the midpoint of a side of the polygon; its length, h, is less than the circle radius. Also, let each side of the polygon have length s; then the sum of the sides, ns, is less than the circle circumference. The polygon area consists of n equal triangles with height h and base s, thus equals ¹⁄₂nhs. But since h < r and ns < c, the polygon area must be less than the triangle area, ¹⁄₂cr, a contradiction. Therefore our supposition that C might be greater than T must be wrong.

[edit] Not less

Circle with square and octagon circumscribed, showing area gap

Suppose the circle area may be less than the triangle area. Let D denote the deficit amount. Circumscribe a square, so that the midpoint of each edge lies on the circle. If the total area gap between the square and the circle, G₄, is greater than D, slice off the corners with circle tangents to make a circumscribed octagon, and continue slicing until the gap area is less than D. The area of the polygon, P_n, must be less than T.

$\begin{align} D &{}= T - C \\ &{}> G_n \\ P_n &{}= C + G_n \\ &{}< C + D \\ P_n &{}< T \end{align}$

This, too, forces a contradiction. For, a perpendicular to the midpoint of each polygon side is a radius, of length r. And since the total side length is greater than the circumference, the polygon consists of n identical triangles with total area greater than T. Again we have a contradiction, so our supposition that C might be less than T must be wrong as well.

Therefore it must be the case that the area of the circle is precisely the same as the area of the triangle. This concludes the proof.

[edit] Reassembly proof

Following Leonardo da Vinci, we can use inscribed regular polygons in a different way. Suppose we inscribe a hexagon. Cut the hexagon into six triangles by splitting it from the center. Two opposite triangles both touch two common diameters; slide them along one so the radial edges are adjacent. They now form a parallelogram, with the hexagon sides making two opposite edges, one of which is the base, s. Two radial edges form slanted sides, and the height is h (as in the Archimedes proof). In fact, we can assemble all the triangles into one big parallelogram by putting successive pairs next to each other. The same is true if we increase to eight sides and so on. For a polygon with 2n sides, the parallelogram will have a base of length 2ns, and a height h. As the number of sides increases, the length of the parallelogram base approaches half the circle circumference, and its height approaches the circle radius. In fact, the parallelogram becomes a rectangle with width πr and height r.

polygon		parallelogram
n	side	base	height	area
4	1.4142136	2.8284271	0.7071068	2.0000000
6	1.0000000	3.0000000	0.8660254	2.5980762
8	0.7653669	3.0614675	0.9238795	2.8284271
10	0.6180340	3.0901699	0.9510565	2.9389263
12	0.5176381	3.1058285	0.9659258	3.0000000
14	0.4450419	3.1152931	0.9749279	3.0371862
16	0.3901806	3.1214452	0.9807853	3.0614675
96	0.0654382	3.1410320	0.9994646	3.1393502

[edit] Onion proof

Circle area ring integration

Using calculus, we can sum the area incrementally, partitioning the disk into thin concentric rings like the layers of an onion. At each thickness, t, the accumulated area is 2πt dt, the circumferential length of the ring times its infinitesimal width. This gives an elementary integral for a disk of radius r.

$\begin{align} \mathrm{Area}(r) &{}= \int_0^{r} 2 \pi t \, dt \\ &{}= \left[ (2\pi) \frac{t^2}{2} \right]_{t=0}^{r}\\ &{}= \pi r^2. \end{align}$

[edit] Fast approximation

For Archimedes, calculations to approximate the area numerically were laborious, and he stopped with a polygon of 96 sides. A faster method uses ideas of Willebrord Snell (Cyclometricus, 1621) followed up by Christiaan Huygens (De Circuli Magnitudine Inventa, 1654), described in Behnke et al. (1983).

Given a circle, let u_n be the perimeter length of an inscribed regular n-gon, and let U_n be the perimeter length of a circumscribed regular n-gon. Then we have the following doubling formulae.

$\begin{align} U_{2n} &{}= \frac{2 U_{n} u_{n}}{ U_{n} + u_{n}} , &\qquad& \text{harmonic mean} \\ u_{2n} &{}= \sqrt{U_{2n} u_{n}} , &\qquad& \text{geometric mean} \end{align}$

Archimedes doubled a hexagon four times to get a 96-gon. For a unit circle, an inscribed hexagon has u₆ = 6, and a circumscribed hexagon has U₆ = 4√3. We have the luxury of decimal notation and our two equations, so we can quickly double seven times:

n	k	u_k	U_k	(u_k + U_k)/4	k = 6×2ⁿ
0	6	6.0000000	6.9282032	3.2320508
1	12	6.2116571	6.4307806	3.1606094
2	24	6.2652572	6.3193199	3.1461443
3	48	6.2787004	6.2921724	3.1427182
4	96	6.2820639	6.2854292	3.1418733
5	192	6.2829049	6.2837461	3.1416628
6	384	6.2831152	6.2833255	3.1416102
7	768	6.2831678	6.2832204	3.1415970

A best rational approximation to the last average is ³⁵⁵⁄₁₁₃, which is an excellent value for π. But Snell proposes (and Huygens proves) a tighter bound than Archimedes.

$k \frac{3 \sin \frac{\pi}{k}}{2+\cos\frac{\pi}{k}} < \pi < k \frac{2 \sin \frac{\pi}{k} + \tan \frac{\pi}{k}}{3}$

Thus we could get the same approximation, with decimal value about 3.14159292, from a 48-gon.

[edit] Derivation

Circle with similar triangles: circumscribed side, inscribed side and complement, inscribed split side and complement

Let one side of an inscribed regular n-gon have length s_n and touch the circle at points A and B. Let A′ be the point opposite A on the circle, so that A′A is a diameter, and A′AB is an inscribed triangle on a diameter. By Thales' theorem, this is a right triangle with right angle at B. Let the length of A′B be c_n, which we call the complement of s_n; thus c_n²+s_n² = (2r)². Let C bisect the arc from A to B, and let C′ be the point opposite C on the circle. Thus the length of CA is s_2n, the length of C′A is c_2n, and C′CA is itself a right triangle on diameter C′C. Because C bisects the arc from A to B, C′C perpendicularly bisects the chord from A to B, say at P. Triangle C′AP is thus a right triangle, and is similar to C′CA since they share the angle at C′. Thus all three corresponding sides are in the same proportion; in particular, we have C′A : C′C = C′P : C′A and AP : C′A = CA : C′C. The center of the circle, O, bisects A′A, so we also have triangle OAP similar to A′AB, with OP half the length of A′B. In terms of side lengths, this gives us

$\begin{align} c_{2n}^2 &{}= \left( r + \frac{1}{2} c_n \right) 2r \\ c_{2n} &{}= \frac{s_n}{s_{2n}} . \end{align}$

In the first equation C′P is C′O+OP, length r+¹⁄₂c_n, and C′C is the diameter, 2r. For a unit circle we have the famous doubling equation of Ludolph van Ceulen,

$c_{2n} = \sqrt{2+c_n} . \,\!$

If we now circumscribe a regular n-gon, with side A″B″ parallel to AB, then OAB and OA″B″ are similar triangles, with A″B″ : AB = OC : OP. Call the circumscribed side S_n; then this is S_n : s_n = 1 : ¹⁄₂c_n. (We have again used that OP is half the length of A′B.) Thus we obtain

$c_n = 2\frac{s_n}{S_n} . \,\!$

Call the inscribed perimeter u_n = ns_n, and the circumscribed perimenter U_n = nS_n. Then combining equations, we have

$c_{2n} = \frac{s_n}{s_{2n}} = 2 \frac{s_{2n}}{S_{2n}} ,$

so that

$u_{2n}^2 = u_n U_{2n} . \,\!$

This gives a geometric mean equation. We can also deduce

$2 \frac{s_{2n}}{S_{2n}} \frac{s_n}{s_{2n}} = 2 + 2 \frac{s_n}{S_n} ,$

$\frac{2}{U_{2n}} = \frac{1}{u_n} + \frac{1}{U_n} .$

This gives a harmonic mean equation.

[edit] Dart approximation

Circle area Monte Carlo integration. Estimate by these 900 samples is 4×⁷⁰⁹⁄₉₀₀ = 3.15111.

When more efficient methods of finding areas are not available, we can resort to “throwing darts”. This Monte Carlo method uses the fact that if random samples are taken uniformly scattered across the surface of a square in which a disk resides, the proportion of samples that hit the disk approximates the ratio of the area of the disk to the area of the square. This should be considered a method of last resort, as it requires enormous numbers of samples to get useful accuracy. (An estimate good to 10⁻ⁿ requires about 100ⁿ random samples.)

[edit] Finite rearrangement

We have seen that by partitioning the disk into an infinite number of pieces we can reassemble the pieces into a rectangle. A remarkable fact discovered relatively recently is that we can dissect the disk into a large but finite number of pieces, and then reassemble the pieces into a square of equal area. This is called Tarski's circle-squaring problem. However, the nature of the proof is such that we cannot say how to construct such a partition, we can only say that it must exist.

[edit] Generalizations

We can stretch a disk to form an ellipse. If the semimajor and semiminor axes of the ellipse are a and b, the area of the ellipse is ab times the area of a unit disk.

We can also consider analogous measurements in higher dimensions. For example, we may wish to find the volume inside a sphere. When we have a formula for the surface area, we can use the same kind of “onion” approach we used for the disk.

[edit] References

Archimedes (2002). "Measurement of a circle", The Works of Archimedes, trans. T. L. Heath, Dover, pp. 91–93. ISBN 978-0-486-42084-4. (Originally published by Cambridge University Press, 1897, based on J. L. Heiberg's Greek version.)
Beckmann, Petr (1976). A History of Pi. St. Martin's Griffin. ISBN 978-0-312-38185-1.
Gerretsen, J.; Verdenduin, P. (1983). "Chapter 8: Polygons and Polyhedra", in H. Behnke, F. Bachmann, K. Fladt, H. Kunle (eds.): Fundamentals of Mathematics, Volume II: Geometry, S. H. Gould (trans.), MIT Press, pp. 243–250. ISBN 0-262-52094-X (ISBN 978-0-262-52094-2).
(Originally Grundzüge der Mathematik, Vandenhoeck & Ruprecht, Göttingen, 1971.)
Laczkovich, Miklós (1990). "Equidecomposability and discrepancy: A solution to Tarski's circle squaring problem". Journal für die reine und angewandte Mathematik (Crelle’s Journal) 404: 77–117. ISSN 0075-4102.